How Do You Calculate Forces on a Fairground Ride Ramp?

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To calculate the forces on a fairground ride ramp, the weight of the car is analyzed, revealing that the component acting parallel to the ramp is approximately 2.3 kN. This force results in a deceleration of about 3.3 m/s² for a fully loaded car with a mass of 690 kg. The car's initial speed of 22 m/s necessitates a ramp length of 73 meters to ensure it stops before reaching the end. The discussion suggests using conservation of energy for part c to simplify calculations. Additional clarification is needed for part d, which was not included in the original query.
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Homework Statement


Hi i have a question I've been doing and need to know if i am completely wrong (or not) any help woul be really appreciated.

A fairground ride ends with the car moving up aramp at a slope of 20degrees to the horizontal

Homework Equations


a)The car carrying its maximum load of passengers has a total weight of 6.8kN.Show that the component of the weight acting parallel to the ramp is about 2.3kN.

b) The mass of the fully loaded car is 690kg. Show that the force in part a will deccelerate the car at about 3.3ms^2

c) The car enters the ramp at 22ms^1.Calculate the minimum length that the ramp must be in order for the car to stop before it reaches the end.Neglect the length of the car.

The Attempt at a Solution



Ok for a) i did tan20degrees =opp/adj so tan20degreesx6.8kn=adj =2.47kN (horizontal component).

b) i did parallel component = f=ma so a=2.3/690 = 3.3ms^2

c) i used v=u^2+2as = 0= (22ms^1)^2 +2(-3.3ms^2x s)
=484=-6.6xs so s =484/6.6 = 73m.

d) i am not to sure about this one can anybody help me with this please?

Thanks,Chris
 
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Part a) asks for the parallel component
c) would be solved easier with conservation of energy.
You did not post question d).
 
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